Kapalci

2022-06-25

The form of the partial fraction decomposition of a rational function is given below.
$\frac{x-3{x}^{2}-26}{\left(x+1\right)\left({x}^{2}+9\right)}=\frac{A}{x+1}+\frac{Bx+C}{{x}^{2}+9}$
What are the values of $A,B$ and $C$?
So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$
I also found the indefinite integral, which equals
$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)-3\mathrm{log}\left(1+x\right)+C$

Expert

Multiplying either sides by $\left(x+1\right)\left({x}^{2}+9\right),$
$x-3{x}^{2}-26=A\left({x}^{2}+9\right)+\left(Bx+C\right)\left(x+1\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-3{x}^{2}+x-26={x}^{2}\left(A+B\right)+x\left(B+C\right)+9A+C$
Compare the constants and the coefficients of $x,{x}^{2}$ of the above identity to form three simultaneous equations with three unknowns A,B,C

George Bray

Expert

Multiply by $x+1$ and let $x\to -1$ to obtain
$\frac{-1-3-26}{\left(-1{\right)}^{2}+9}=A\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}A=-3.$
Multiply both sides by $x$, let $x\to \mathrm{\infty }$, and use $A=-3$ to obtain
$-3=A+B\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}B=0.$
Finally, evaluate at $x=0$, and use $A=-27/9$:
$\frac{-26}{9}=A+\frac{C}{9}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}C=9\left(-\frac{26}{9}+\frac{27}{9}\right)=1.$

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